View Single Post
 ebaines Posts: 10,055, Reputation: 5539 Expert #2 Jul 7, 2008, 02:38 PM
Start by breaking the problem in two - consider the components of the forces in the horizontal direction seprarately from the components of the forces in the vertical direction. You want to have $Sigma F_x = 0$ and $Sigma F_y = 800N$. The 600N force that goes up to the left has a horizontal component of 600N * (-4/5) (note that its sign is negative as it goes to the left), and a vertical component of 600N * 3/5. The 400N force going up to the right has a horizontal component of 400N * cos(30), and a vertical componenet of 400* sin(30). The unknown force $F_1$ has a horizontal component of $F_1 * cos( heta)$ and a vertical component of $F_1 sin( heta)$. So summing all the horizontal forces and setting that equal to 0 yields:

$
Sigma F_x = 600(-4/5) + F_1 cos( heta) + 400N * cos(30) = 0
$

From this you can determine the value for $F_1 cos( heta )$.

Then sum the forces in the vertical direction:

$
Sigma F_y = 600*3/5 +F_1 sin( heta ) + 400*sin(30) = 800
$

This gives you $F_1 sin( heta )$. The value of $F_1$ can then be found from:

$
F_1 = sqrt { (F_1 cos ( heta ))^2 + (F_1 sin ( heta ))^2 }
$

And the direction of $F_1$ is:

$
heta = Atan ( frac {F_1 sin ( heta ) } {F_1 cos( heta ) }
$